Question

The temperature and pressure at Simla are 15.0°C and 72.0 cm of mercury and at Kalka these are 35.0°C and 76.0 cm of mercury. Find the ratio of air density at Kalka to the air density at Simla.

Use R=8.314J K^-1 mol^-1

Answer 1

Here,
Temperature in Simla, T₁= 15 + 273 = 288 K       
Pressure in Simla, P_​1 = 0.72  m of Hg 
Temperature in Kalka, T₂= 35+273 = 308 K
Pressure in Kalka, P_​2 = 0.76  m of Hg 
Let density of air at Simla and Kalka be ρ_{1 and} ρ_{2 respectively. Then,}

\[PV = \frac{m}{M}RT\] 

\[ \Rightarrow \frac{m}{V} = \frac{PM}{RT}\] 

\[ \Rightarrow \rho = \frac{PM}{RT}\]

Thus, \[\rho_1  = \frac{P_1 M}{R T_1}\]

\[\rho_2  = \frac{P_2 M}{R T_2}\]

Taking ratios , we get 

\[\frac{\rho_1}{\rho_2} = \frac{P_1}{T_1} \times \frac{T_2}{P_2}\] 

\[ \Rightarrow \frac{\rho_1}{\rho_2} = \frac{0 . 72}{288} \times \frac{308}{0 . 76}\] 

\[ \Rightarrow \frac{\rho_2}{\rho_1} = 0 . 987\]